 Bivariate Limits - Maple Help

 Bivariate Limits

The limit command has been enhanced for the case of limits of bivariate rational functions with non-isolated singularities. Many such limits that could not be determined previously are now computable. If the limit exists in such a situation, it is either $+\mathrm{∞}$ or $-\mathrm{∞}$. Maple can also determine if the limit does not exist, and then returns $\mathrm{undefined}$.

In Maple 18, all the following limit calls would return unevaluated, but they can now be computed in Maple 2015.

 >
 > $\mathrm{limit}\left(f,\left\{x=0,y=0\right\}\right)$
 ${\mathrm{undefined}}$ (1)
 >
 > $\mathrm{limit}\left(g,\left\{x=0,y=0\right\}\right)$
 ${\mathrm{undefined}}$ (2)
 > $h≔\frac{{x}^{4}-{x}^{2}-{y}^{2}}{{\left(x-y\right)}^{4}}:$
 > $\mathrm{limit}\left(h,\left\{x=0,y=0\right\}\right)$
 ${-}{\mathrm{\infty }}$ (3)

Let us plot these three functions in the neighborhood of the origin.

In the first example, $f$  tends to $+\mathrm{∞}$ on one side of the singularity $y=-x$ and to $-\mathrm{∞}$ on the other side (as shown in the following plot). Therefore, the limit at the origin does not exist.

 >
 >
 > $\mathrm{plots}:-\mathrm{display}\left(\mathrm{pf1},\mathrm{pf2}\right)$ Now, consider the second example.

 > $s≔\mathrm{plots}:-\mathrm{spacecurve}\left(\left[x,-x,-1\right],x=-0.1..0.1,\mathrm{color}=\mathrm{red},\mathrm{thickness}=3\right):\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{pg}≔\mathrm{plot3d}\left(g,x=-0.1..0.1,y=-0.1..0.1,\mathrm{axes}=\mathrm{boxed},\mathrm{view}=-10..100,\mathrm{numpoints}=40000\right):\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{plots}:-\mathrm{display}\left(s,\mathrm{pg}\right)$ $g$ tends to $+\mathrm{∞}$ close to the singularity $y=x$. However, along the anti-diagonal $y=-x$, the limit is finite:

 > $\mathrm{eval}\left(g,y=-x\right)$
 ${-1}$ (4)

Thus, $g$ does not have a limit at the origin. In fact, any number$\ge -1$ can occur, namely, as the limit along the ray  for $-1\le a<1$:

 >
 $\frac{{4}{{x}}^{{2}}{a}}{{\left({-}{a}{x}{+}{x}\right)}^{{2}}}$ (5)
 > $\mathrm{limit}\left(,x=0\right)$
 $\frac{{4}{a}}{{\left({a}{-}{1}\right)}^{{2}}}$ (6)
 > $\mathrm{plot}\left(,a=-3..1.1,\mathrm{view}=-2..10\right)$ In the last example, $h$ tends to $-\mathrm{∞}$ on both sides of the singularity $y=x$.

 > $\mathrm{plot3d}\left(h,x=-0.1..0.1,y=-0.1..0.1,\mathrm{axes}=\mathrm{boxed},\mathrm{view}=-100000..1\right)$ However, in this case the limit along any ray  with $a\ne 1$ is $-\mathrm{∞}$ as well.

 >
 $\frac{{-}{{a}}^{{2}}{{x}}^{{2}}{+}{{x}}^{{4}}{-}{{x}}^{{2}}}{{\left({-}{a}{x}{+}{x}\right)}^{{4}}}$ (7)
 > $\mathrm{limit}\left(,x=0\right)$
 ${-}{\mathrm{signum}}{}\left(\frac{{{a}}^{{2}}{+}{1}}{{\left({a}{-}{1}\right)}^{{4}}}\right){\mathrm{\infty }}$ (8)
 > $\mathrm{factor}\left(\right)$
 ${-}{\mathrm{signum}}{}\left(\frac{{{a}}^{{2}}{+}{1}}{{\left({a}{-}{1}\right)}^{{4}}}\right){\mathrm{\infty }}$ (9)
 >
 ${-}{\mathrm{\infty }}$
 ${-}{\mathrm{\infty }}$ (10)

You can prove that the limit exists and is $-\mathrm{∞}$ for any curve approaching the origin by using the theory of Lagrange multipliers. The extremal values (maxima and minima) of the function $h$  on the circle, for a fixed radius $r$, satisfy the condition that the gradient of the function and the gradient of the constraint equation of the circle are parallel:

 > $C≔{x}^{2}+{y}^{2}-{r}^{2}:$
 > $\mathrm{with}\left(\mathrm{VectorCalculus}\right):$
 > $\mathrm{df}≔\mathrm{normal}~\left(\mathrm{Jacobian}\left(\left[h\right],\left[x,y\right]\right)\right)$
 $\left[\begin{array}{cc}-\frac{2\left(2{x}^{3}y-{x}^{2}-xy-2{y}^{2}\right)}{{\left(x-y\right)}^{5}}& \frac{2\left(2{x}^{4}-2{x}^{2}-xy-{y}^{2}\right)}{{\left(x-y\right)}^{5}}\end{array}\right]$ (11)
 > $\mathrm{dC}≔\mathrm{Jacobian}\left(\left[C\right],\left[x,y\right]\right)$
 $\left[\begin{array}{cc}2x& 2y\end{array}\right]$ (12)
 > $\mathrm{eq}≔\mathrm{factor}\left(\mathrm{numer}\left(\mathrm{normal}\left({\mathrm{df}}_{1,1}{\mathrm{dC}}_{1,2}-{\mathrm{df}}_{1,2}{\mathrm{dC}}_{1,1}\right)\right)\right)$
 ${\mathrm{eq}}{≔}{-}{8}\left({{x}}^{{3}}{-}{x}{-}{y}\right)\left({{x}}^{{2}}{+}{{y}}^{{2}}\right)$ (13)

Thus, the local maximum and minimum values of $f$  on $C$ occur when both $C=0$ and $\mathrm{eq}=0$. For the bivariate limit, this means that it is sufficient to consider only those critical paths satisfying $\mathrm{eq}=0$. However, you also need to consider the global suprema and infima, which may occur close to the singularity $y=x$.
In the example, the factor ${x}^{2}+{y}^{2}$ of $\mathrm{eq}$ does not admit any real paths, so there is only one critical path given by ${x}^{3}-x-y=0$ (or, equivalently, $y=-x+{x}^{3}$).

 > $\mathrm{normal}\left(\mathrm{eval}\left(h,y=-x+{x}^{3}\right)\right)$
 ${-}\frac{{{x}}^{{2}}{-}{1}}{{{x}}^{{2}}{\left({{x}}^{{2}}{-}{2}\right)}^{{3}}}$ (14)
 > $\mathrm{limit}\left(,x=0\right)$
 ${-}{\mathrm{\infty }}$ (15)

In order to certify the limit close to the singularity $y=x$ as well, you cannot take the limit along the singularity. Instead, consider two curves that approach the singularity closely from the top and from the bottom, respectively:

 > $\mathrm{c1}≔y=x+{x}^{2}$
 ${\mathrm{c1}}{≔}{y}{=}{{x}}^{{2}}{+}{x}$ (16)
 > $\mathrm{c2}≔y=x-{x}^{2}$
 ${\mathrm{c2}}{≔}{y}{=}{-}{{x}}^{{2}}{+}{x}$ (17)
 > > $\mathrm{normal}\left(\mathrm{eval}\left(h,\mathrm{c1}\right)\right)$
 ${-}\frac{{2}\left({x}{+}{1}\right)}{{{x}}^{{6}}}$ (18)
 > $\mathrm{limit}\left(,x=0\right)$
 ${-}{\mathrm{\infty }}$ (19)
 > $\mathrm{normal}\left(\mathrm{eval}\left(h,\mathrm{c2}\right)\right)$
 $\frac{{2}\left({x}{-}{1}\right)}{{{x}}^{{6}}}$ (20)
 > $\mathrm{limit}\left(,x=0\right)$
 ${-}{\mathrm{\infty }}$ (21)
 > See Also